Answer:
Explanation:
Brownian motion is a random (irregular) motion of particles e.g smoke particle. The set up in the diagram can be used to observe the motion of smoke.
1. The apparatus used are:
A is a source of light
B is a converging lens
C is a glass smoke cell
D is a microscope
2. The uses of the apparatus are:
A - produces the light required to so as to see clearly the movement of the particles.
B - converges the rays of light from the source to the smoke cell.
C - is made of glass and used for encamping the smoke particles so as not to mix with air.
D - is used for the clear view or observation or study of the motion of the smoke particles in the cell.
This effect is explained by increased chain entanglements at higher molecular weights. Increasing the degree of crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Deformation by drawing increases the tensile strength of a semicrystalline polymer.
The pressure at a certain depth underwater is:
P = ρgh
P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth
The pressure exerted on the submarine window is:
P = F/A
P = pressure, F = force, A = area
The area of the circular submarine window is:
A = π(d/2)²
A = area, d = diameter
Set the expressions for the pressure equal to each other:
F/A = ρgh
Substitute A:
F/(π(d/2)²) = ρgh
Isolate h:
h = F/(ρgπ(d/2)²)
Given values:
F = 1.1×10⁶N
ρ = 1030kg/m³ (pulled from a Google search)
g = 9.81m/s²
d = 30×10⁻²m
Plug in and solve for h:
h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)
h = 1540m
Answer:
![\Delta \theta = 56\,rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%2056%5C%2Crad)
![\Delta \theta \approx 3208.564^{\circ}](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%5Capprox%203208.564%5E%7B%5Ccirc%7D)
Explanation:
El ángulo barrido en el intervalo de tiempo dado es (The covered angle in the given time interval is):
![\Delta \theta = \omega \cdot \Delta t](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%5Comega%20%5Ccdot%20%5CDelta%20t)
![\Delta \theta = \left(14\,\frac{rad}{s} \rjght)\cdot (4\,s)](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%20%5Cleft%2814%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%5Crjght%29%5Ccdot%20%284%5C%2Cs%29)
![\Delta \theta = 56\,rad](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%3D%2056%5C%2Crad)
![\Delta \theta \approx 3208.564^{\circ}](https://tex.z-dn.net/?f=%5CDelta%20%5Ctheta%20%5Capprox%203208.564%5E%7B%5Ccirc%7D)
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